3.68.35 \(\int \frac {1}{5} e^{-6 x/5} (-75-18 x+e^{2 x} (25+4 x)) \, dx\)

Optimal. Leaf size=22 \[ e^{-x/5} \left (3 e^{-x}+e^x\right ) (5+x) \]

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Rubi [B]  time = 0.09, antiderivative size = 47, normalized size of antiderivative = 2.14, number of steps used = 8, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 6742, 2194, 2176} \begin {gather*} 3 e^{-6 x/5} x+15 e^{-6 x/5}-\frac {5}{4} e^{4 x/5}+\frac {1}{4} e^{4 x/5} (4 x+25) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-75 - 18*x + E^(2*x)*(25 + 4*x))/(5*E^((6*x)/5)),x]

[Out]

15/E^((6*x)/5) - (5*E^((4*x)/5))/4 + (3*x)/E^((6*x)/5) + (E^((4*x)/5)*(25 + 4*x))/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int e^{-6 x/5} \left (-75-18 x+e^{2 x} (25+4 x)\right ) \, dx\\ &=\frac {1}{5} \int \left (-75 e^{-6 x/5}-18 e^{-6 x/5} x+e^{4 x/5} (25+4 x)\right ) \, dx\\ &=\frac {1}{5} \int e^{4 x/5} (25+4 x) \, dx-\frac {18}{5} \int e^{-6 x/5} x \, dx-15 \int e^{-6 x/5} \, dx\\ &=\frac {25}{2} e^{-6 x/5}+3 e^{-6 x/5} x+\frac {1}{4} e^{4 x/5} (25+4 x)-3 \int e^{-6 x/5} \, dx-\int e^{4 x/5} \, dx\\ &=15 e^{-6 x/5}-\frac {5}{4} e^{4 x/5}+3 e^{-6 x/5} x+\frac {1}{4} e^{4 x/5} (25+4 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 0.82 \begin {gather*} e^{-6 x/5} \left (3+e^{2 x}\right ) (5+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-75 - 18*x + E^(2*x)*(25 + 4*x))/(5*E^((6*x)/5)),x]

[Out]

((3 + E^(2*x))*(5 + x))/E^((6*x)/5)

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fricas [A]  time = 0.50, size = 17, normalized size = 0.77 \begin {gather*} {\left (3 \, {\left (x + 5\right )} e^{\left (-2 \, x\right )} + x + 5\right )} e^{\left (\frac {4}{5} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((4*x+25)*exp(x)^2-18*x-75)/exp(1/5*x)/exp(x),x, algorithm="fricas")

[Out]

(3*(x + 5)*e^(-2*x) + x + 5)*e^(4/5*x)

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giac [A]  time = 0.19, size = 18, normalized size = 0.82 \begin {gather*} {\left (x + 5\right )} e^{\left (\frac {4}{5} \, x\right )} + 3 \, {\left (x + 5\right )} e^{\left (-\frac {6}{5} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((4*x+25)*exp(x)^2-18*x-75)/exp(1/5*x)/exp(x),x, algorithm="giac")

[Out]

(x + 5)*e^(4/5*x) + 3*(x + 5)*e^(-6/5*x)

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maple [A]  time = 0.05, size = 24, normalized size = 1.09

method	result	size
risch	\(\frac {\left (25+5 x \right ) {\mathrm e}^{\frac {4 x}{5}}}{5}+\frac {\left (15 x +75\right ) {\mathrm e}^{-\frac {6 x}{5}}}{5}\)	\(24\)
default	\(15 \,{\mathrm e}^{-\frac {6 x}{5}}+5 \,{\mathrm e}^{\frac {4 x}{5}}+3 x \,{\mathrm e}^{-\frac {6 x}{5}}+x \,{\mathrm e}^{\frac {4 x}{5}}\)	\(27\)
norman	\(\left (15+x \,{\mathrm e}^{2 x}+3 x +5 \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-\frac {6 x}{5}}\)	\(29\)
meijerg	\(-15+\frac {25 \,{\mathrm e}^{-\frac {6 x}{5}}}{2}+\frac {5 \,{\mathrm e}^{2 x} {\mathrm e}^{-2 x} \left (1-\frac {\left (2-\frac {8 x}{5}\right ) {\mathrm e}^{\frac {4 x}{5}}}{2}\right )}{4}-\frac {25 \,{\mathrm e}^{2 x} {\mathrm e}^{-2 x} \left (1-{\mathrm e}^{\frac {4 x}{5}}\right )}{4}+\frac {5 \left (2+\frac {12 x}{5}\right ) {\mathrm e}^{-\frac {6 x}{5}}}{4}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((4*x+25)*exp(x)^2-18*x-75)/exp(1/5*x)/exp(x),x,method=_RETURNVERBOSE)

[Out]

1/5*(25+5*x)*exp(4/5*x)+1/5*(15*x+75)*exp(-6/5*x)

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maxima [B]  time = 0.38, size = 35, normalized size = 1.59 \begin {gather*} \frac {1}{4} \, {\left (4 \, x - 5\right )} e^{\left (\frac {4}{5} \, x\right )} + \frac {1}{2} \, {\left (6 \, x + 5\right )} e^{\left (-\frac {6}{5} \, x\right )} + \frac {25}{4} \, e^{\left (\frac {4}{5} \, x\right )} + \frac {25}{2} \, e^{\left (-\frac {6}{5} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((4*x+25)*exp(x)^2-18*x-75)/exp(1/5*x)/exp(x),x, algorithm="maxima")

[Out]

1/4*(4*x - 5)*e^(4/5*x) + 1/2*(6*x + 5)*e^(-6/5*x) + 25/4*e^(4/5*x) + 25/2*e^(-6/5*x)

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mupad [B]  time = 4.05, size = 14, normalized size = 0.64 \begin {gather*} {\mathrm {e}}^{-\frac {6\,x}{5}}\,\left ({\mathrm {e}}^{2\,x}+3\right )\,\left (x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-x)*exp(-x/5)*((18*x)/5 - (exp(2*x)*(4*x + 25))/5 + 15),x)

[Out]

exp(-(6*x)/5)*(exp(2*x) + 3)*(x + 5)

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sympy [A]  time = 0.15, size = 20, normalized size = 0.91 \begin {gather*} \left (x + 5\right ) e^{\frac {4 x}{5}} + \left (3 x + 15\right ) e^{- \frac {6 x}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((4*x+25)*exp(x)**2-18*x-75)/exp(1/5*x)/exp(x),x)

[Out]

(x + 5)*exp(4*x/5) + (3*x + 15)*exp(-6*x/5)

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